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Solution Manual For Introductory Circuit Analysis, 12th Edition

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    Chapter 1 1 Chapter 1 1.  2.  3.  4.  = t d =                   h 1 min 60 min 1 s 60 ft 5,280 mi 1 s 10 ft 000 , 20 = 1363.64 mph 5. 4 min       min 60 h 1 = 0.067 h  = h 1.067 mi 31  t d = 29.05 mph 6. 100 yds             ft 5,280 mi 1 yd 1 ft 3 = 0.0568 mi 60 mi 1 h 1 min h 60 min 60 s             = 0.0167 mi/s t = 0.0568 mi 0.0167 mi/s d   = 3.40 s 7. a.                   s 60 min 1 min 60 h 1 mi ft 280 , 5 h mi 95 = 139.33 ft/s b. t = ft/s 139.33 ft 60   d = 0.431 s c.  =                    ft 5,280 mi 1 h 1 min 60 min 1 s 60 s 1 ft 60 t d = 40.91 mph 8.  9.  10.  11. MKS, CGS,  C = ) 36 ( 9 5 ) 32 68 ( 9 5 ) 32 F ( 9 5      = 20  SI: K = 273.15 +  C = 273.15 + 20 = 293.15 12.       J 1 lb - ft 0.7378 J 1000 = 737.8 ft-lbs

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    Chapter 1 2 13. 0.5 yd                   in. 1 cm 54 . 2 ft 1 in. 12 yd 1 ft 3 = 45.72 cm 14. a. °F = 2(°C) + 30° = 40° + 30° = 70° b. °F = 9 9 ( C) 32 (20 ) 32 5 5        = 68° c. very close d. 30°C  90°F vs. 86° 5°C  40°F vs 41° 15. a. 14.6 b. 56.0 c. 1,046.1 d. 1 16 = 0.0625 = 0.1 e. 3.14159 = 3.1 16. a. 14.60 b. 56.04 c. 1,046.06 d. 1 16 = 0.0625 = 0.06 e. 3.14159 = 3.14 17. a. 14.603 b. 56.042 c. 1,046.060 d. 1 16 = 0.0625 = 0.063 e. 3.14159 = 3.142 18. a. 10 4 b. 10 6 c. 10 3 d. 10  3 e. 10 0 f. 10  1 19. a. 15  10 3 b. 5  10  3 c. 2.4  10 6 d. 60  10 3 e. 4.02  10  4 f . 2  10  10 20. a. 4.2  10 3 + 48.0  10 3 = 52.2  10 3 = 5.22  10 4 b. 90  10 3 + 360  10 3 = 450  10 3 = 4.50  10 5 c. 50  10  5  6  10  5 = 44  10  5 = 4.4  10  4 d. 1.2  10 3 + 0.05  10 3  0.4  10 3 = 0.85  10 3 = 850 21. a. (10 2 )(10 3 ) = 10 5 = 100  10 3 b. (10  2 )(10 3 ) = 10 1 = 10 c. (10 3 )(10 6 ) = 1  10 9 d. (10 2 )(10  5 ) = 1  10  3 e. (10  6 )(10  10 6 ) = 10 f. (10 4 )(10  8 )(10 28 ) = 1  10 24

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    Chapter 1 3 22. a. (50  10 3 )(2  10  3 ) = 100  10 0 = 100 b. (2.2  10 3 )(2  10  3 ) = 4.4  10 0 = 4.4 c. (82  10  6 )(1.2  10  6 ) = 98.4 d. (30  10  4 )(4  10  3 )(7  10 8 ) = 840  10 1 = 8.40  10 3 23. a. 10 2 /10 4 = 10  2 = 10  10  3 b. 10  2 /10 3 = 10  5 = 10  10  6 c. 10 4 /10  3 = 10 7 = 10  10 6 d. 10  7 /10 2 = 1.0  10  9 e. 10 38 /10  4 = 1.0  10 42 f. 2 10 / 100  = 10 1 /10  2 = 1  10 3 24. a. (2  10 3 )/(8  10  5 ) = 0.25  10 8 = 2.50  10 7 b. (4  10  3 )/(4  10 6 ) = 4/4  10  9 = 1  10  9 c. (22  10  5 )/(5  10  5 ) = 22/5  10 0 = 4.4 d. (78  10 18 )/(4  10  6 ) = 1.95  10 25 25. a. (10 2 ) 3 = 1.0  10 6 b. (10  4 ) 1/2 = 10.0  10  3 c. (10 4 ) 8 = 100.0  10 30 d. (10  7 ) 9 = 1.0  10  63 26. a. (2  10 2 ) 2 = 4  10 4 b. (5  10  3 ) 3 = 125  10  9 c. (4  10  3 )(3  10  3 ) 2 = (4  10  3 )(9  10 4 ) = 36  10 1 = 360 d. ((2  10  3 )(0.8  10 4 )(0.003  10 5 )) 3 = (4.8  10 3 ) 3 = (4.8) 3  (10 3 ) 3 = 110.6  10 9 = 1.11  10 11 27. a. (  10  3 ) 2 = 1.0  10  6 b. 3 4 2 10 ) 10 )( 10 (  = 10  2 /10 3 = 1.0  10  5 c. 4 4 4 2 6 4 2 2 3 10 10 10 ) 10 )( 10 ( 10 ) 10 ( ) 10 (      = 1.0  10  8 d. 4 4 3 10 ) 10 )( 10 (  = 10 7 /10  4 = 1.0  10 11 e. (1  10  4 ) 3 (10 2 )/10 6 = (10  12 )(10 2 )/10 6 = 10  10 /10 6 = 1.0  10  16 f.       10 1 ) 10 )( 10 ( 1 10 ) 10 ( ) 10 )( 10 ( 3 4 3 2 2 3 2 2       = 1.0  10  1 28. a. 4 2 2 2 10 3 ) 10 ( ) 10 3 (   = (9  10 4 )(10 2 )/(3  10 4 ) = (9  10 6 )/(3  10 4 ) = 3  10 2 = 300 b. 3 8 3 2 4 10 8 10 16 ) 20 ( ) 10 4 (     = 2  10 5 c. 4 8 2 2 2 4 10 4 10 36 ) 10 2 ( ) 10 6 (        = 9.0  10 12

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    Chapter 1 4 d. 5 2 5 3 / 1 6 10 2 10 3 10 2 ) 10 27 (        = 1.5  10  7 = 150.0  10  9 e. 4 8 4 2 6 4 2 2 3 10 2 10 48 10 2 ) 10 3 )( 10 16 ( 10 2 ) 10 3 ( ) 10 4 (              = 24.0  10 12 f. (16  10  6 ) 1/2 (10 5 ) 5 (2  10  2 ) = (4  10  3 )(10 25 )(2  10  2 ) = 8  10 20 = 800.0  10 18 g. 2 1/ 2 3 3 2 2 4 5 2 (3 10 ) 1.60 10 (2 10 )(8 10 ) (7 10 )                     9 4 2 1/ 2 10 5 1 6 10 10 (27 10 )(2.56 10 )(16 10 ) 49 10 (69.12 10 )(4 10 ) 276.48 10 49 10 49 10                     = 5.64  10 4 = 56.4  10 3 29. Scientific: a. 2.05 × 10 1 b. 5.04 × 10 4 c. 6.74 × 10  4 d. 4.60 × 10  2 Engineering: a. 20.46 × 10 0 b. 50.42 × 10 3 c. 674.00 × 10  6 d. 46.00 × 10  3 30. Scientific a. 5.0 × 10  2 b. 4.5 × 10 1 c. 1/32 = 0.03125 = 3.125 × 10  2 d. 3.14159 = 3.142 × 10 0 Engineering: a. 50.0 × 10  3 b. 0.045 × 10 3 c. 31.25 × 10  3 d. 3.142 × 10 0

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    Chapter 1 5 +2 31. a. 6  10 4 = 0.06  10 +6 = 0.06 × 10 +6  3  3 b. 0.4  10  3 = 400  10  6 = 400 × 10  6 +3  2 +3 +3 c. 50  10 5 = 5000  10 3 = 5  10 6 = 0.005  10 9 = 0.005 × 10 9 +2  3  3 +4  3  3 d. 12  10  7 = 0.0012  10  3 = 1.2  10  6 = 1200  10  9 = 1200 × 10 9  4 +3 +3  3 32. a. 0.05  10 0 s = 50  10  3 s = 50 ms +3 +3 b. 2000  10  6 s = 2  10  3 s = 2 ms  3  3 c. 0.04  10  3 s = 40  10  6 s = 40  s +3 +6 d. 8400  10  12 s  0.0084  10  6 s = 0.0084  s  6

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    Chapter 1 6 +3 increase by 3 10 0 e. 100  10 3  10  3 m = 0.1  10 3 m = 0.1 km  3 33. a. 1.5 min       min 1 s 60 = 90 s b. 2 × 10  2 h             min 1 s 60 h 1 min 60 = 72 s c. 0.05 s        s 10 s 1 6  = 0.05  10 6  s = 50  10 3  s d. 0.16 m        m 10 mm 1 3 = 0.16  10 3 mm = 160 mm e. 1.2  10  7 s        s 10 ns 1 9 = 1.2  10 2 ns = 120 ns f. 4  10 8 s                   h 24 day 1 min 60 h 1 s 60 min 1 = 4629.63 days 34. a. 80  10  3 m       m 1 cm 100 = 8000  10  3 cm = 8 cm b. 60 cm             m 1000 km 1 cm 100 m 1 = 60  10  5 km c. 12 × 10  3 m 6 1 m 10 m         = 12  10  3 × 10 +6  m = 12 × 10 3  m d. 60 cm 2 1 m 1 m 100 cm 100 cm             = 60  10  4 m 2 35. a. 100 in.       in. 39.37 m 1 = 2.54 m b. 4 ft             in. 39.37 m 1 ft 1 in. 12 = 1.22 m

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    Chapter 1 7 c. 6 lb       lb 1 N 45 . 4 = 26.7 N d. 60  10 3 dynes             N 4.45 lb 1 dynes 10 N 1 5 = 0.13 lb e. 150,000 cm             in. 12 ft 1 cm 2.54 in. 1 = 4921.26 ft f. 0.002 mi                   in. 39.37 m 1 ft 1 in. 12 mi 1 ft 5280 = 3.22 m 36. 5280 ft , 5280 ft       ft 3 yd 1 = 1760 yds 5280 ft             in. 39.37 m 1 ft 1 in. 12 = 1609.35 m, 1.61 km 37. 60 mi 5280 ft 12 in. 1 m 1 h 1 min h 1 mi 1 ft 39.37 in. 60 min 60 s = 26.82 m/s 38. 10 km                         ft 5280 mi 1 in. 12 ft 1 m 1 in. 39.37 km 1 m 1000 = 6.214 mi  = min 6.5 mi 1 , t = min 6.5 mi 1 mi 6.214   d = 40.39 min 39. 100 yds             ft 1 in. 12 yd 1 ft 3 = 3600 in  3600 quarters 40. 60 mph: t = 500 mi 60 mi/h d   = 8.33 h = 8 h:19.8 min 75 mph: t = 500 mi 75 mi/h d   = 6.67 h = 6h:40.2 min difference = 1h:28.6 minutes 41. d =  t =                           cm 100 m 1 min 1 s 60 h 1 min 60 h 016 . 0 s cm 600 = 345.6 m

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    Chapter 1 8 42. d = 86 stories                   ft 12 9 step 1 story ft 14 = 1605 steps  =           seconds 60 minute 1 seconds 5 . 802 second steps 2 steps 1605  d t t d = 13.38 minutes 43. d = (86 stories)              ft 5,280 mile 1 ft 1204 story ft 14 = 0.228 miles min 10.22 min mile 0.228 miles  = 44.82 min/mile 44. minute ft 1056 1mile ft 5,280 min 5 mile 1 mile min 5         , distance = 86 stories       story ft 14 = 1204 ft  = min ft 1056 ft 1204     d t t d = 1.14 minutes 45. a.       J 1054.35 Btu 1 J 5 = 4.74  10  3 Btu b. 24 ounces             gallons 172 . 264 m 1 ounces 128 gallon 1 3 = 7.1  10  4 m 3 c. 1.4 days       day 1 s 400 , 86 = 1.21  10 5 s d. 1 m 3             gallon 1 pints 8 m 1 gallons 264.172 3 = 2113.38 pints 46. 6(4 × 2 + 8) = 96 47. (42 + 6/5)/3 = 14.4 48. 2 2 2 5 3  = 5.044 49. MODE = DEGREES: cos 21.87  = 0.928 50. MODE = DEGREES: tan  1 (3/4) = 36.87  51.   2 400 /(6 10 / 5)  = 7.071

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    Chapter 1 9 52. 205  10  6 53. 1.20  10 12 54. 6.667  10 6 + 0.5  10 6 = 7.17  10 6

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    Chapter 2 10 Chapter 2 1.  2. a. F = 2 9 2 2 1 m) (1 C) C)(2 1 )( 10 9 (   r Q Q k = 18  10 9 N b. F = k 2 2 1 r Q Q =   2 9 m) 3 ( C 2 C) 1 )( 10 9 (  = 2  10 9 N c. F = 2 9 2 2 1 m) (10 C) C)(2 1 )( 10 9 (   r Q Q k = 0.18  10 9 N d. Exponentially, 3 1 10 m 1 m r r  = 10 while 9 1 9 2 18 10 N 0.18 10 N F F    = 100 3. a. r = 1 mi: 1 mi                   in. 37 . 39 m 1 ft 1 in. 12 mi 1 ft 5280 = 1609.35 m F = 9 6 -6 3 1 2 2 2 6 (9 10 )(8 10 C)(40 10 C) 2880 10 (1609.35 m) 2.59 10 kQ Q r          = 1.11  N b. r =10 ft: 12 in. 1 m 10 ft 1 ft 39.37 in.             = 3.05 m F = 3 3 1 2 2 2 2880 10 2880 10 9.30 (3.05 m) kQ Q r       = 0.31 N c.       in. 39.37 m 1 16 in. 1 = 1.59 mm F = 3 3 3 1 2 2 3 2 6 2880 10 2880 10 1138.34 10 (1.59 10 m) 2.53 10 kQ Q r             N = 1138.34 kN 4.  5. Q 1 = Q 2 = Q ; 2 2 2 2 2 1 1 1 1 1 2 2 2 2 1 2 2 ; F r F r kQ kQ k F Q F k k r r r            and 2 F  2 1 1 2 2 r F r 6. F = 4 2 6 9 2 1 2 2 1 10 6 . 3 ) 10 20 )( 10 9 (        F Q kQ r r Q kQ = 10 mm

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    Chapter 2 11 7. F = 2 1 2 2 1 2 2 1 m) 2 ( 8 . 1 Q kQ Q kQ r Q kQ    = 4(1.8) = 7.2 a. F = 2 2 2 1 ) 10 ( 2 . 7  r Q kQ = 72 mN b. Q 1 / Q 2 = 1/2  Q 2 = 2 Q 1 7.2 = kQ 1 Q 2 = (9  10 9 )( Q 1 )(2 Q 1 ) = 9  10 9   2 1 2 Q 9 1 2 1 9 10 18 2 . 7 10 18 2 . 7      Q Q = 20  C Q 2 = 2 Q 1 = 2(2  10  5 C) = 40  C 8. V = 1.2 J 20 C W Q   = 60 kV 9. W = VQ = (60 V)(8 mC) = 0.48 J 10. Q = 120 J 20 mV W V   = 6 mC 18 6.242 10 electrons 1 C        = 37.45 × 10 15 electrons 11. Q = V 9 J 72  V W = 8 C 12. a. W = QV = (1 × 10 12 electrons)(40 V) = 40 × 10 12 eV b. 40 × 10 12 eV 18 1 C 6.242 10 electrons        = 6.41 μ J 13. I = 12 mC 2.8 s Q t  = 4.29 mA 14. I = 312 C (2)(60 s) Q t  = 2.60 A 15. Q = It = (40 mA)(0.8)(60 s) = 1.92 C 16. Q = It = (250 mA)(1.2)(60 s) = 18.0 C 17. t = mA 2 mC 6  I Q = 3 s 18. 21.847  10 18 electrons        electrons 10 6.242 C 1 18 = 3.5 C I = s 12 C 3.5  t Q = 0.29 A

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    Chapter 2 12 19. Q = It = (4 mA)(90 s) = 360 mC 360 mC        C 1 electrons 10 242 . 6 18 = 2.25  10 18 electrons 20. I = s) (1.2)(60 C 86  t Q = 1.194 A > 1 A (yes) 21. 0.84  10 16 electrons        electrons 10 6.242 C 1 18 = 1.346 mC I = ms 60 mC 346 . 1  t Q = 22.43 mA 22. a. Q = It = (2 mA)(0.01  s) = 2  10  11 C 2  10  11 C 18 6.242 10 electrons 1 ¢ 1 C electron              = 1.25  10 8 ¢ = $1.25  10 6 = 1.25 million b. Q = It = (100  A)(1.5 ns) = 1.5  10  13 C 1.5  10  13 C              electron 1 $ C 1 electrons 10 242 . 6 18 = 0.94 million (a) > (b) 23. Q = It = (200  10  3 A)(30 s) = 6 C V = C 6 J 40  Q W = 6.67 V 24. Q = It = min) 5 . 0 ( min C 420       = 210 C V = C 210 J 742  Q W = 3.53 V 25. Q = V 24 J 0.4  V W = 0.0167 C I = 3 0.0167 C 5 10 s Q t    = 3.34 A 26. I = h 40 Ah 200 (hours) rating Ah  t = 5 A 27. Ah = (0.8 A)(75 h) = 60.0 Ah 28. t (hours) = A 1.28 Ah 32 rating Ah  I = 25 h

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    Chapter 2 13 29. 40 Ah(for 1 h): W 1 = VQ = V  I  t = (12 V)(40 A)(1 h)             min 1 s 60 h 1 min 60 = 1.728  10 6 J 60 Ah(for 1 h): W 2 = (12 V)(60 A)(1 h)             min 1 s 60 h 1 min 60 = 2.592  10 6 J Ratio W 2 / W 1 = 1.5 or 50% more energy available with 60 Ah rating. For 60 s discharge: 40 Ah = It = I               min 60 h 1 s 60 min 1 s 60 = I (16.67  10  3 h) and I = h 10 16.67 Ah 40 3 -  = 2400 A 60 Ah = It = I               min 60 h 1 s 60 min 1 s 60 = I (16.67  10  3 h) and I = h 10 16.67 Ah 60 3 -  = 3600 A I 2 / I 1 = 1.5 or 50 % more starting current available at 60 Ah 30. 0.75(18 Ah) = 13.5 Ah   250 mA 31. (18 Ah  15.5 Ah)/18 Ah × 100% = 13.89% 32. At 100 mA, discharge time  120 H; At 25 mA, discharge time  425 h;  300 h more at 25 mA 33. I = 3 Ah 6.0 h = 500 mA Q = It = (500 mA)(6 h)             min 1 s 60 h 1 min 60 = 10.80 kC W = QV = (10.8 kC)(12 V)  129.6 kJ 34.  35.  36.  37. a. 0.5 in 2.54 cm 1 in       = 1.27 cm 1.27 cm 30 kV cm       = 38.1 kV b. 1.27 cm 270 kV cm       = 342.9 kV c. 342.9 kV:38.1 kV = 9:1

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    Chapter 2 14 38.  39.  40.  41. 

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    Chapter 3 15 Chapter 3 1. a. 0.5 in. = 500 mils b. 0.02 in.       in. 1 mils 1000 = 20 mils c. 4 1 in. = 0.25 in.       in. 1 mils 1000 = 250 mils d. 10 mm = 10 × 10  3 m 39.37 in 1000 mils 1 m 1 in             = 393.7 mils e. 0.01 ft             in. 1 mils 10 ft 1 in. 12 3 = 120 mils f. 0.1 cm             in. 1 mils 1000 cm 2.54 in. 1 = 39.37 mils 2. a. A CM = (30 mils) 2 = 900 CM b. 0.016 in. = 16 mils, A CM = (16 mils) 2 = 256 CM c. 1 8 = 0.125" = 125 mils, A CM = (125 mils) 2 = 15.63  10 3 CM d. 1 cm             in. 1 mils 1000 cm 2.54 in. 1 = 393.7 mils, A CM = (393.7 mils) 2 = 155  10 3 CM e. 0.02 ft             in. 1 mils 1000 ft 1 in. 12 = 240 mils, A CM = (240 mils) 2 = 57.60  10 3 CM f. 4 × 10  3 m 39.37 in. 1000 mils 1 m 1 in             = 157.48 mils , A CM = (157.48 mils) 2 = 24.8 × 10 3 CM 3. A CM = ( d mils ) 2  d mils = CM A a. d = CM 1600 = 40 mils = 0.04 in. b. d = 820 CM = 28.64 mils = 0.029 in. c. d = CM 40,000 = 200 mils = 0.2 in. "

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